Đáp án:

Bạn tham khảo lời giải ở dưới nhé!!!

Giải thích các bước giải:

18,

\(\begin{array}{l}
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{Zn}} = 0,1mol\\
 \to {n_{{H_2}}} = {n_{Zn}} = 0,1mol\\
 \to {n_{{H_2}}} = \dfrac{{PV}}{{RT}}\\
1b{\rm{ar}} = 0,99atm\\
 \to {V_{{H_2}}} = \dfrac{{nRT}}{P} = \dfrac{{0,1 \times 0,082 \times (273 + 25)}}{{0,99}} = 2,468l
\end{array}\)

22,

\(\begin{array}{l}
F{\rm{e}} + 2HCl \to F{\rm{e}}C{l_2} + {H_2}\\
{n_{HCl}} = 0,2mol\\
{n_{{H_2}}} = \dfrac{{PV}}{{RT}} = \dfrac{{0,99 \times 4,958}}{{0,082 \times (273 + 250)}} = 0,1145mol\\
{n_{{H_2}}} = \dfrac{1}{2}{n_{HCl}} = 0,1mol < 0,1145mol\\
 \to {n_{F{\rm{e}}}} = \dfrac{1}{2}{n_{HCl}} = 0,1mol\\
 \to {m_{F{\rm{e}}}} = 5,6g
\end{array}\)

24,

\(\begin{array}{l}
CaO + 2HCl \to CaC{l_2} + {H_2}O\\
{n_{CaO}} = 0,4mol\\
 \to {n_{CaC{l_2}}} = {n_{CaO}} = 0,4mol\\
 \to m = {m_{CaC{l_2}}} = 44,4g
\end{array}\)

25,

\(\begin{array}{l}
2K{\rm{O}}H + {H_2}S{O_4} \to {K_2}S{O_4} + 2{H_2}O\\
{n_{K{\rm{O}}H}} = 0,2mol\\
 \to {n_{{H_2}S{O_4}}} = \dfrac{1}{2}{n_{K{\rm{O}}H}} = 0,1mol\\
 \to m = {m_{{H_2}S{O_4}}} = 9,8g
\end{array}\)